In the figure, calculate the measure of the radius of the circle inscribed in the triangle $IMN$, where $a$ and $b$ are arrows and $c$ is the radius of the semicircumference .
Remarks: $O$ is $I$ are respectively circumcenter and incenter of triangle $ABC$.
(S:$\frac{\sqrt{abc}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$)
I try
$\triangle OMN: MN^2 = c^2+c^2 \implies MN =c\sqrt2$
$\angle BAM \cong \angle BCM \cong \angle\cong \angle MNA$
$\angle BAN \cong \angle BAC \cong \angle ANO \cong \angle CMN$
$ \triangle MIN \sim \triangle AIC(A.A.)\implies \frac{c\sqrt2}{c}=\sqrt2 = k$
$AG = GC$
$AF = BF$
Best Answer
Let us enumerate different "facts" :
$$(c-a)^2+(c-b)^2=c^2 \tag{1}$$
$$xp=S\tag{2}$$
between its inradius $x$, its half-perimeter $p$ and its area $S$.
Explanation for triangle $AMI$ : $\angle AMI=90°$ because it subtends a diameter and $\angle MAI=45°=90°/2$ because it subtends arc MBN which is subtended by the center angle $\angle MON$. (similar proof for triangle $CNI$).
As a consequence
$$AM=IM=\sqrt{2ac} \ \text{and} \ CN=IN=\sqrt{2bc}\tag{3}$$
Let us prove the second relationship in (3) (it is the same reasoning for the first relationship).
Pythagoras theorem in triangle $CGN$ gives, using (1)
$$CN^2=b^2+(c-a)^2=b^2+c^2-(c-b)^2=2bc$$
As a consequence of (3), the half perimeter of triangle $IMN$ is :
$$p=\frac12 \left(\sqrt{2ac} + \sqrt{2bc}+ c \sqrt{2} \right)=\sqrt{c}\frac{\sqrt{2}}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right)\tag{4}$$
Besides, it is rather easy to prove that angle $\angle MIN = 135°$ giving, for the area of triangle $MIN$ :
$$S=\frac12 IM.IN \sin 135°=\frac12 \sqrt{2ac}\sqrt{2bc} \frac{\sqrt{2}}{2}=c\sqrt{ab}\frac{\sqrt{2}}{2}\tag{5}$$
Now, we can use (2) with expressions coming from (3) and (5) :
$$x=\frac{S}{p}=\frac{c\sqrt{ab}\frac{\sqrt{2}}{2}}{\sqrt{c}\frac{\sqrt{2}}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right)}=\frac{\sqrt{abc}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$$